Question

Prove that:
$cose{c}^6\theta -{\cot }^6\theta =1+3cose{c}^2\theta .{\cot }^2\theta$

Answer 1

L.H.S.
$=cose{c}^6\theta -{\cot }^6\theta$
$=(cose{c}^2\theta {)}^3-({\cot }^2\theta {)}^3$
$=(cose{c}^2\theta -{\cot }^2\theta )\left[\right(cose{c}^2\theta {)}^2+({\cot }^2\theta {)}^2+cose{c}^2.\theta {\cot }^2\theta ]$
$[\because cose{c}^2\theta -{\cot }^2\theta =1]$
$=1[cose{c}^4\theta +{\cot }^4\theta +cose{c}^2\theta .{\cot }^2\theta ]$
$=\left[\right(cose{c}^2\theta -{\cot }^2\theta {)}^2+3cose{c}^2\theta .{\cot }^2\theta ]$
$=1+3cose{c}^2\theta .{\cot }^2\theta$ = R.H.S.
Hence Proved.